Question

Area of the rectangle formed by the ends of latus rectums of the ellipse $25x^2+4y^2=100$ is

• A. $\frac{16\sqrt{21}}{5}$
• B. $\frac{32\sqrt{21}}{5}$
• C. $\frac{8\sqrt{21}}{5}$
• D. $\frac{7\sqrt{21}}{5}$

Answer 1

The correct option is A $\frac{16\sqrt{21}}{5}$

Given ellipse $25x^2+4y^2=100$

This can be re-written as $\frac{x^2}{4}+\frac{y^2}{25}=1$

Comparing it to the general form of ellipse we get $a=2,b=5$

Here $b>a$

Now eccentricity e is given by $e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{4}{25}}=\frac{\sqrt{21}}{5}$

Now the rectangle is formed by the ends of the latus rectum.

The ends of latus rectum are given by

$\{(\frac{a^2}{b},be),(-\frac{a^2}{b},be),(-\frac{a^2}{b},-be),(\frac{a^2}{b},-be)\}$

Putting values of $a,b,e$ we gethe ends of latus rectum to be

$\{(\frac{4}{5},\sqrt{21}),(-\frac{4}{5},\sqrt{21}),(-\frac{4}{5},-\sqrt{21}),(\frac{4}{5},-\sqrt{21})\}$

Now to find the area of rectangle, find distances between the two points where area $A=l\times b$

$\Rightarrow l=\sqrt{{\left(\sqrt{21}\right)}^2}=\left(2\sqrt{21}\right),b=\sqrt{{\left(\frac{8}{5}\right)}^2}=\frac{8}{5}$

Thus Area $\Rightarrow A=\left(2\sqrt{21}\right)\times \frac{8}{5}=16\frac{\sqrt{21}}{5}$